An object with a mass of $12 k g$ $12 kg$ is on a surface with a kinetic friction coefficient of $1$ $1$ . How much force is necessary to accelerate the object horizontally at $14 \frac{m}{s^{2}}$ $14 m/s^2$ ?

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An object with a mass of $12 k g$ $12 kg$ is on a surface with a kinetic friction coefficient of $1$ $1$ . How much force is necessary to accelerate the object horizontally at $14 \frac{m}{s^{2}}$ $14 m/s^2$ ?

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Answer 1 · 2016-04-20T00:29:07

$290 N [forward]$ $290N["forward"]$

Explanation:

For this problem it is important to note that there are two forces acting on the object in the $x$ $x$ direction, the applied and friction forces.

Start by listing out the given and required values.

$m = 12 k g i, i μ_{k} = 1 i, i a = 14 \frac{m}{s^{2}} i, i F_{app} = ?$ $m=12kgcolor(white)(i)color(teal),color(white)(i)mu_k=1color(white)(i)color(teal),color(white)(i)a=14m/s^2color(white)(i)color(teal),color(white)(i)F_"app"=?$

A free-body diagram can represent the situation as:

![http://www.physicsclassroom.com/class/newtlaws/Lesson-2/Drawing-Free-Body-Diagrams]()

In this case, set the positive directions to be forward and down.

Using $F_{net, y} = m a_{y}$ $F_("net",y)=ma_y$ , add the forces in the $y$ $y$ direction.

$F_{net, y} = m a_{y}$ $F_("net",y)=ma_y$

$F_{N} + F g = m a_{y}$ $F_N+Fg=ma_y$

Since the object does not accelerate vertically, the acceleration in the $y$ $y$ direction is $0 \frac{m}{s^{2}}$ $0m/s^2$ .

$F_{N} + F g = m (0 \frac{m}{s^{2}})$ $F_N+Fg=m(0m/s^2)$

$F_{N} + F g = 0 N$ $F_N+Fg=0N$

$F_{N} = - F g$ $F_N=-Fg$

$color(blue)(|bar(ul(color(white)(a/a)color(black)(F_N=-mg)color(white)(a/a)|)))$

After adding the forces in the $y$ direction, do the same for the forces in the $x$ direction. Rearrange for $F_"app"$ to find the force needed to accelerate the object horizontally.

$F_("net",x)=ma_x$

$F_"app"+F_(f,k)=ma_x$

$F_"app"=ma_x-F_(f,k)$

$F_"app"=ma_x-mu_kF_N$

$F_"app"=ma_x-mu_k(color(blue)(-mg))$

$F_"app"=ma_x+mu_kmg$

Substitute your known values.

$F_"app"=(12kg)(14m/s^2)+(1)(12kg)(9.81m/s^2)$

$F_"app"=285.72N$

$F_"app"~~color(green)(|bar(ul(color(white)(a/a)290Ncolor(white)(a/a)|)))$

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An object with a mass of $12 k g$ $12 kg$ is on a surface with a kinetic friction coefficient of $1$ $1$ . How much force is necessary to accelerate the object horizontally at $14 \frac{m}{s^{2}}$ $14 m/s^2$ ?

1 Answer

Explanation:

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An object with a mass of 12 kg12kg12 kg is on a surface with a kinetic friction coefficient of 1 1 1 . How much force is necessary to accelerate the object horizontally at 14 m/s^214ms2 14 m/s^2?

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Explanation:

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An object with a mass of $12 k g$ $12 kg$ is on a surface with a kinetic friction coefficient of $1$ $1$ . How much force is necessary to accelerate the object horizontally at $14 \frac{m}{s^{2}}$ $14 m/s^2$ ?